Problem: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 17}{x - 4} = \dfrac{10x - 41}{x - 4}$
Answer: Multiply both sides by $x - 4$ $ \dfrac{x^2 - 17}{x - 4} (x - 4) = \dfrac{10x - 41}{x - 4} (x - 4)$ $ x^2 - 17 = 10x - 41$ Subtract $10x - 41$ from both sides: $ x^2 - 17 - (10x - 41) = 10x - 41 - (10x - 41)$ $ x^2 - 17 - 10x + 41 = 0$ $ x^2 + 24 - 10x = 0$ Factor the expression: $ (x - 6)(x - 4) = 0$ Therefore $x = 6$ or $x = 4$ However, the original expression is undefined when $x = 4$. Therefore, the only solution is $x = 6$.